Isomers and tartaric acid

[btcboss2022]

I think something wrong with graph.
If you use L+-tartaric acid(cheaper and easier to find)the D enantiomer(amph,Meth) will be in the solution not in the solid as there appears.
If you use D-tartaric acid(more expensive and difficult to find) the D enantiomer(amph,Meth) will be in the solid.
Thanks.

 

[G.Patton]

His scheme is correct. Salt is precipitated into the solid state. No difference, d-amphetamine l-tartaric acid salt or l-amphetamine d-tartaric salt. You can change acid and get difference isomer in the solid state with following procedures.

P.S. same for methamphetamine.

 

[diogenes]

G.Patton is correct. He basically uses the same method at the end of his procedure, which is used in this topic, he precipitates the D-Amphetamine with excess d-tartaric (aka L-(+)-tartaric acid or the natural form) in the end. This way it forms D-Amphetamine-d-bitartrate aka D-Amphetamine-L-(+)-bitartrate salt. The same principle is used, except he first gets rid of most of the L-Amphetamine isomer by half-molar tartaric acid, which leads to the formation of Amphetamine tartrate. As it happens L-Amphetamine-d-tartrate is much less soluble in alcohols than D-Amphetamine-d- (or L-(+)-) tartrate.

I think the L-Amphetamine-d-bitartrate is not even a stable molecule and it gets oxidized by the excess tartaric acid, this is why the solution gets red. So Patton`s procedure is more elegant (but more labour intensive) and is capable of separating the L-amphetamine which in this topic`s method simply gets destroyed.

 

[Joker_55555]

Hello, does methamphetamine synthesized from the method of ephedrine and pseudoephedrine give the d-methamphetamine isomer with 99% purity? If not, how much?

 

[Sciencenutz]

Yes it does only p2p gives 50/50

 

[Osmosis Vanderwaal]

I'd fact check that. There's other precursors who are chiral

 

[Osmosis Vanderwaal]

I know you guys aren't dumping your L- meth when you can do an inversion and make it D- meth. I've seen it done a couple of ways.

 

[G.Patton]

Hi, yes, it can be done via thiols. These compounds extremely bad smell.

 

[Tweaker]

Have either of you performed this?

 

[G.Patton]

No, I've read about this way.

 

[btcboss2022]

Ok I have the results of my tests trying to follow the video method but the similar reagents that I have on hand and that I usually use.
I divided 70gr of racemic freebase(not steam distilled and no acid-base washing made just extracted and dried from the NaBH4 P2P reduction) in 2 equal parts of 35gr each. With the first 35 gr I did video process but with some different reagents and with the other 35gr I did the method that I use to make that is similar to the video process:

- Method 1(Video method)


- 35gr pure freebase
- 42gr L-tartaric
- 205ml Methanol
The amounts are like video ratio but methanol I decided to use a 12% less due ethanol in video is a 88% solution.
I added the acid to the flask later the methanol and finally the freebase.
Once I started stirring immediately happened what usually happened to me in the rest of the processes I have tried for this purpose the solid salts precipitated quickly forming a mass.
I reflux it during one hour, with heating the mass come a thick liquid.
After the refluxing hour I keep it at RT around 4-5 hours until no new solids are visible created in the mixture.
I filtered it and keep in different pots the solid and the liquid part.
-Liquid part
I get cold the liquid part in the freezer with a NaOH solution to avoid excess heat in the alkalinization step.
I added slowly the NaOH solution to the mixture until ph13 and add it to a separation funnel during 30 min.
I take out the aqueous bottom layer and keep it , the thin oil top layer is kept in a pot.
I extract with DCM the aqueous layer and add these DCM extractions to oil layer.
I dried it with anhydrous sodium sulfate and filtered it.
I evaporated the DCM and 10gr of D-meth freebase obtained(don't looks as clean as it should)
- Solid part
I added more or less the same ml of warm water than gr of the solids weight, they has been dissolved it with stirring and let it cool down.
I added slowly the cold NaOH solution until ph13 and add it to a separation funnel during 30 min.
The next steps are exactly the same as the liquid part process.
After DCM is evaporated I got 15gr of L-meth freebase quite "dirty" too

- Method 2(the one I usually do and similar to the video method)


- 35gr pure freebase
- 41,2gr L-tartaric
- 412ml Methanol
Methanol and tartaric are mixed until complete dissolution, freebase is added while stirring and continue stirring vigorously 2-3min.
Let it at RT during 24 hours until the complete solid mass is formed(if in a couple of hours no solids are formed you should stir it again)
It's filtered and now are exactly the same steps as the other method.

- In the liquid part once I had all the DCM extractions I decided to make an acid-base washing, even the DCM color looks cleaner than the other one, to avoid to be "dirty" like the first method happened and I obtained 10gr of D-meth absolutely clean and pure(pic link at the bottom)
- In the solid part 13gr of enough clean L-meth freebase without acid-base washing obtained.
- Cleaning issue could be fixed during the crystallization stage too are many ways.

Obviously is very difficult to check how much D or L are in reality in the stuff only with Chiral analysis not easy accessible.
I also know that it is not the most correct to compare two methods using some different reagents but in reality they do practically the same function in the process.

In short, similar methods and yields the main differences are that in the second one you can avoid the reflux(so could be easier) and the final freebases looks cleaner than the others method.

When I will have free time again like this weekend(unfortunately not usual :-( only will be remaining to test in small scale the"mexican"method for meth isomer separation is quite different and I only did it in big scale although I finally decided to use the method explained for big scale too.
These would be the 3 methods that I did and I can talk about them with experience.

About when and how to make an acid-base washing instead steam distillation...etc etc needs a complete and long post the same for crystallization.

I'm thinking to post all the possible and more profitable options during the route from BMK(5449) to D-Meth HCL but this would be a long work and I don't know when I will have enough time again hahah at least is planned.

Thanks.

 

[Acab1312]

Thank you for writing down your method, do they do the isomeric separation only once or do they repeat it?

 

[btcboss2022]

I did it only once.

 

[Davidrobinson]

Fb is miscible with methanol, water is miscible with methanol , dcm is miscible with methanol .
So how is it possible to extract?
Unless you are using a lot of water

 

[btcboss2022]

I don't understand your question so much.

 

[Davidrobinson]

In the part what’s says “liquid part” your add naoh ,to the methanol which has the salt in, yes, to liberate the FB. How are you extracting the FB?
How is there a aqueous layer, when water is miscible with methanol? And how is there a upper oil layer when FB is soluble in methanol?
Am just asking because this has me confused .
Thanks

 

[btcboss2022]

Ok now is more clear, I'm not chemist so possibly I'm not the right person to answer with the chemical literature that you apply but I can explain how I did it in that case and how I do it in all similar processes.
Until I know(I could be wrong obviously but it worked in that way in the 100% of the cases that I did) in acid mixtures(PH2) the freebases(ONLY FREEBASE NOT P2P FOR EXAMPLE) remains out of water or aqueous part and it that case won't be solved by any solvent for this reason solvents are used to wash the freebase in acid mixtures and these solvents are discarded instead to keep it as is supposed to do with an organic solvent.
In alkali mixtures(PH12-13) freebases are "liberated"that means it's in water or aqueous side, like freebase and water has different densities the 2 layers are formed but some freebase could remains in water for this reason the aqueous layer is extracted with an organic solvent that in alkali mixtures solves the organic(freebase).

I don't know if that info(that could be corrected by someone if not accurate) answer your question in the way that you need but I hope at least it help you.
Maybe a Doctor in chemistry or a chemist with degree could help you more sorry.

 

[Davidrobinson]

Ye just simple chemistry really isn’t it, don’t really need a doctor to answer it . Maybe it’s temperature , maybe it wouldn’t work so well on bigger scale. Could be a lot of fb still in the methanol

 

[btcboss2022]

I guess not but I will check it, next time after extraction I will evaporate all the methanol and extract the mixture again to see if any FB is remaining ok?

 

[btcboss2022]

I made a simple test I extracted water-methanol solution later I evaporated methanol and extracted again.
There is a practically undetectable amount of FB in the second extraction Im not sure if it was remaining in methanol or in the water.
Anyway there are 2 options to fix that:
- When you turns the solution alkali you can separate oil layer, evaporate the methanol from the mixture and extract the mixture with DCM or directly turn it alkali with a NaOH methanol solution and evaporate all methanol.

 

[btcboss2022]

This is the way how isomer separation looks like after 1 hour with method 2 ;-)

 

[Acab1312]

Hello btcboss2022 I tested their method, unfortunately it didn't work for me. Everything worked out wonderfully with the precipitation. Unfortunately, with the separation of the freebase from methanol I did you something wrong, could you explain the separation of the liquid phase again in how far you separate it?

 

[btcboss2022]

Is not my method hahah but ok I understood you, what you did exactly and what happened?

 

[Tweaker]

What's the 3rd method? I'm assuming it's much more suited to large scale manufacturing?

 

[btcboss2022]

1 mol tartaric in 200ml of H2O dissolve and add 1 mol of freebase stirring.
Leave it 24H RT.
Filter....

 

[xxxx-8888]

I followed method 2, why doesn't the solid appear?

 

[Osmosis Vanderwaal]

I stayed up all night combing through scholar and patents and here is my findings

L(+) tartaric acid (2R,3R) is the naturally occuring isomer.Since it's chiraliity ir R,R, it bends light and is optically active. D(-)(2S,3S) is also optically active. These 2 compounds are enantiomers. They share the same boiling, freezing, temperature, mol.wt. ect.
The diastereomer of both is meso tartaric acid. (2R,3S or 2S,3R) the 2 mesos are enantiomers.
What makes a compound optically active is bending light. What make it bend light? Having an uneven plane of carbon substituants. For instance L(+)tartaric acid is active. It bends light right and right ( I say right but I am just using right as a reference). Same with D(-) tartaric acid, it bends light " left left". Meso tartaric acid is chiral but not optically active. Because (for instance) 2R,3s bends light right the left again for a net gain of 0°.
The reason we are talking about optically active compounds is because active compounds have a higher melting point and a higher boiling point. Another way to look at it is they solidify(crystallize) at a higher temperature. Yes an optically active compound crystallizes before an inactive one. Now we get to the point.
If you use D(ss) tartaric acid the D-(S) Methamphetamine is optically active (S,S,S) and the L(R) Methamphetamine is not, (R,R,S)
The opposite is also true. If you use the natural isomer of tartaric acid, the levo-methamphetamine crystallizes first, so if you use natural tartaric acid, the crystals are the trash. The rework. The mother liquor contains the Dextro-meth.
If you use racemic D,L-tartaric acid on racemic meth, you crystallize racemic Methamphetamine tartrate. Wasted your time.
Summary of the invention; use one tartaric acid isomer and if you want the D-crystals, use the D tartaric acid. Otherwise use the L and filter the crystals out.
Bring your references when you come to debate these facts.

 

[btcboss2022]

You mean that video process is wrong then? I never used racemic tartaric honestly so I can't talk about it but with Dextro tartaric acid isomer separation is done I can't know exactly in what % but it works.

 

[Osmosis Vanderwaal]

I'll have
To watch the video again. If it is contradictory to my statement then it must be incorrect. Yes, as I stated, the D isomer of tartaric acid will precipitate the D-meth first. The % depends upon the % of enantiomeric excess you achieved in the reduction reaction. 50% is the theoretical model but 40-60% is a reasonable expectation I believe. There's a law that relates to yield who's name I forget but it deals with equilibrium and basically says you would have to neutralize the rm and reconjugate it 3-6 times to achieve a maximum separation. It's the same law that has one to extract a solvent several times and combine them